Theorem ddif 4071

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif	⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3436	. . . . 5 ⊢ 𝑥 ∈ V
2		eldif 3897	. . . . 5 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴))
3	1, 2	mpbiran 706	. . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
4	3	con2bii 358	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
5	1	biantrur 531	. . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
6	4, 5	bitr2i 275	. 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
7	6	difeqri 4059	1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   ∧ wa 396   = wceq 1539   ∈ wcel 2106  Vcvv 3432   ∖ cdif 3884

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890

This theorem is referenced by:  complss  4081  dfun3  4199  dfin3  4200  invdif  4202  ssindif0  4397  difdifdir  4422